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Lesson 26: The Divergence Theorem August 3rd, 2015 Section 16.9 At the end of our lecture on Green’s Theorem, we derived two vector versions of it. One involved the curl and the other divergence. The divergence version is Z C F · n ds = ZZ div F(x , y ) dA D where C is the positively oriented boundary curve of the plane region D. What we will see today is that this integral essentially extends to vector fields on R3 . Section 16.9 Theorem (The Divergence Theorem) Let E be a simple solid region and let S be the boundary surface of E , given with positive (outward) orientation. Let F be a vector field whose component functions have continuous partial derivatives on an open region that contains E . Then ZZ F · dS = S ZZZ div F dV E Here E is a simple region (i.e a region of type I, II, and III such as a sphere, ellipsoid, or cube) and S is a closed surface whose unit normal points outward from E . The theorem is true for more general regions E as well. Section 16.9 Example Verify that the Divergence Theorem is true for the vector field F(x , y , z) = 3x i + xy j + 2xz k on the region E which is the cube bounded by the planes x = 0, x = 1, y = 0, y = 1, z = 0, and z = 1. Section 16.9 Example Find the (outward) flux of the vector field F(x , y , z) = z i + y j + x k over the unit sphere x 2 + y 2 + z 2 = 1. Section 16.9 Example Evaluate RR S F · dS where 2 F(x , y , z) = xy i + (y 2 + e xz ) j + sin(xy ) k and S is the surface of the region E bounded by the parabolic cylinder z = 1 − x 2 and the planes z = 0, y = 0, and y +z =2 Section 16.9 When E is a region lying between two surfaces S1 and S2 with S1 lying inside S2 , we can still apply the Divergence Theorem. Consider the picture below So S1 is the sphere, S2 is the cube (ignore the fact that it’s not smooth), and E is the region between them. Let n1 be the unit normal of S1 and n2 the unit normal of S2 . Then the boundary of E is S1 ∪ S2 , and the unit normal n of E is n = n2 on S2 and n = −n1 on S1 . Section 16.9 Applying the Divergence Theorem, this gives us ZZZ div F dV = ZZ E F · dS = ZZ S = ZZ S1 =− F · n dS S F · (−n1 ) dS + ZZ F · dS + ZZ S1 S2 F · n2 dS F · dS S2 Section 16.9 ZZ Example Recall that when an electric charge Q is positioned at the origin, the electric field created is E(x) = εQ x |x|3 where x = hx , y , zi is a position vector. Use the Divergence Theorem to show that the electric flux of E through any closed surface S that encloses the origin is ZZ E · dS = 4πεQ S Section 16.9 As we’ve seen, divergence (like curl) also relates to fluid flow. Let v(x , y , z) be the velocity field of a fluid with constant density ρ. Then F = ρv is the rate of flow per unit area. Take a point P0 (x0 , y0 , z0 ) in the fluid and consider a ball Ba centered at P0 of radius a (with a very small). Since div F is continuous, div F(P) ≈ div F(P0 ) in Ba . The flux over the boundary sphere Sa is then approximately ZZ ZZZ ZZZ F · dS = Sa div F dV ≈ Ba div F(P0 ) dV = div F(P0 )V (Ba ) Ba Letting a → 0 gives a better approximation and suggests that 1 ZZ div F(P0 ) = lim F · dS a→0 V (Ba ) Sa Section 16.9 This last equation tells us that div F(P0 ) is the net rate of outward flux per unit volume at P0 (hence the name divergence). If div F(P) > 0, the net flow is outward near P. In this case we call P a source. If div F(P) < 0, the net flow is inward near P and P is called a sink. Section 16.9 Example Use the Divergence Theorem to calculate S F · dS, where F(x , y , z) = x 2 sin y i + x cos y j − xz sin y k and S is the “fat sphere” x 8 + y 8 + z 8 = 8. RR Section 16.9 Example Compute the flux of F(x , y , z) = |r|r across S, where r = x√i + y j + z k and S consists of the hemisphere z = 1 − x 2 − y 2 and the disc x 2 + y 2 ≤ 1 in the xy -plane. Section 16.9